Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 ~V . The de-Broglie wavelengths are λ_e and λ_p for electrons and protons respectively. The ratio

Correct answer: D. (λ_e)/(λ_p)=2 (m_p)/(m_e) Energy in joule (E) = charge × potential diff. in volt E_ electron =q_ c V and E_ proton =q_ p 4 ~V de-Broglie wavelength λ= h P = h 2 mE array r λ_ e = h 2 ~m_ e eV and λ_ P = h 2 ~m_ e e 4 ~V ( q_ e =q_ P ) array aligned λ_ e λ_ P & = h 2 ~m_ e eV h 2 ~m_ P e 4 ~V = 2 ~m_ P e 4 ~V 2 ~m_ e eV & =2 m_ P m_ e aligned

JEE Main 201323 AprilOnline

Exam: JEE Main
Year: 2013
Date: 23 April
Mode: Online
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2013 · 23 April · Online · Physics · Dual Nature of Matter

Electrons are accelerated through a potential difference V and...

Electrons are accelerated through a potential difference

\mathrm{V}

and protons are accelerated through a potential difference

4 \mathrm{~V}

. The de-Broglie wavelengths are

\lambda_e

and

\lambda_p

for electrons and protons respectively. The ratio of

\frac{\lambda_e}{\lambda_p}

is given by: (given

m_e

is mass of electron and

m_p

is mass of proton).

Options

A $\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_p}{m_e}}$
B $\frac{\lambda_e}{\lambda_p}=\sqrt{\frac{m_e}{m_p}}$
C $\frac{\lambda_e}{\lambda_p}=\frac{1}{2} \sqrt{\frac{m_e}{m_p}}$
D $\frac{\lambda_e}{\lambda_p}=2 \sqrt{\frac{m_p}{m_e}}$

Correct answer

D. $\frac{\lambda_e}{\lambda_p}=2 \sqrt{\frac{m_p}{m_e}}$

Step-by-step solution

Energy in joule (E) = charge

\times

potential diff. in volt

\mathrm{E}_{\text{electron }}=\mathrm{q}_{\mathrm{c}} \mathrm{V}

and

\mathrm{E}_{\text{proton }}=\mathrm{q}_{\mathrm{p}} 4 \mathrm{~V}

de-Broglie wavelength

\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}

\begin{array}{r} \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}} \text{ and } \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{e} 4 \mathrm{~V}}} \\ \left(\because \mathrm{q}_{\mathrm{e}}=\mathrm{q}_{\mathrm{P}}\right) \end{array}

\begin{aligned} \therefore \frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{P}}} & =\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{P}} \mathrm{e} 4 \mathrm{~V}}}}=\sqrt{\frac{2 \mathrm{~m}_{\mathrm{P}} \mathrm{e} 4 \mathrm{~V}}{2 \mathrm{~m}_{\mathrm{e}} \mathrm{eV}}} \\ & =2 \sqrt{\frac{\mathrm{m}_{\mathrm{P}}}{\mathrm{m}_{\mathrm{e}}}}\end{aligned}

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